ch9.3: binary search a number from a sorted & twisted array

最新推荐文章于 2024-02-19 17:01:40 发布

原创最新推荐文章于 2024-02-19 17:01:40 发布 · 702 阅读

0 ·

CC 4.0 BY-SA版权

ctci 专栏收录该内容

34 篇文章

订阅专栏

本文介绍了一种O(logn)复杂度的算法，在已排序但被未知次数旋转的数组中查找特定元素，适用于数据结构和算法领域的学习者。

Given a sorted array of n integers that has been rotated an unknown number of times, give an O(log n) algorithm that finds an element in the array. You may assume that the array was originally sorted in increasing order.

if(x<= a[low] && x>a[mid]) // must have = for low because the a[mid] cannot be equal but a[low] maybe

This code doesn't work for duplicate case.

#include <iostream>

int search(int a[], int low, int high, int x){
    while(low <= high){
        int mid = (low+high)>>1;
        if(x==a[mid]) return mid;
        if(a[low]> a[mid]){
            if(x<= a[low] && x>a[mid]) // must have = for low
                low = mid+1;
            else
                high = mid -1;
        }
        else{
            if(x<a[mid] && x>=a[low])  // must have = for low
                high = mid -1;
            else
                low = mid + 1;
        }    
    }
    return -1;
}


using namespace std;

int main()
{
   cout << "Hello World" << endl; 
   int a[12] = {
        15, 16, 19, 20, 25, 1, 3, 4, 5, 7, 10, 14
    };
    int b[19] = {
        2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,2
    };
    cout<<search(a, 0, 11, 4)<<endl;
    cout<<search(b, 0, 18, 3)<<endl;
    return 0;
}